# Let $G$ be a finite group of order $n$ and let $a_1,…,a_n$ be elements of $G$ (not necessarily distinct).

Let $G$ be a finite group of order $n$ and let $a_1,...,a_n$ be elements of $G$ (not necessarily distinct). Show that there exists integers $p,q$ with $1 \leq p \leq \ q \leq n$ such that $(a_p)(a_{p+1}) \dots(a_q)=1$. I have no idea no how to start this question. Anyone can help ?

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## 1 Answer

Let $P_1=a_1 \,;\, P_2=a_1a_2,.. \,;\, P_q=a_1...a_q \,;....\,; P_n=a_1a_2..a_n$.

If some $P_i$ is $e$ you are done. Otherwise, they can take only $n-1$ values, so by the pigeon hole principle there exists some $p < q$ so that $P_p =P_q$.

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actually what motivates you to consider $P_n=a_1 a_2 ... a_n$? Usually the problem that I encounter is that I don know what to consider. –  Idonknow Feb 2 '13 at 21:13
@Idonknow I don't know, experience? :) The idea is that they are products of consecutive terms, so you need to look at products of consecutive. Starting at 1 is natural since you have to start somewhere, and you can't get too much information just by looking at one, so I looked at all. –  N. S. Feb 2 '13 at 21:31
I edited to change all but the first $P_1$ to $P_i$ in the two other places it appears –  Barbara Osofsky Feb 2 '13 at 21:56
@BarbaraOsofsky ty. –  N. S. Feb 2 '13 at 21:57
@N.S.: I don understand why 'Otherwise, they can take only $n−1$ values ' . 'they' refers to what ? –  Idonknow Feb 3 '13 at 3:54